Time, Speed and Distance

Let the usual speed be v and usual time be t. New speed is 0.8v. Since distance is constant, v * t = 0.8v * (t + 15). Dividing by v, t = 0.8t + 12. Thus, 0.2t = 12, so t = 60 minutes.

Let distance be D. Time at 3 km/h = D/3. Time at 4 km/h = D/4. Difference = 20 min + 10 min = 30 min = 0.5 hours. D/3 - D/4 = 0.5. D/12 = 0.5. D = 6 km.

Let d be distance, t be scheduled time. d/24 = t + 5/60. New speed = 24 * 1.25 = 30 kmph. d/30 = t - 4/60. Subtract equations: d/24 - d/30 = 9/60. d(5-4)/120 = 9/60. d/120 = 9/60 => d = 18 km.

Let distance be D. Time taken = 3 hours. Speed 1 = 2 km/h, Speed 2 = 3 km/h. Average speed = 2*2*3 / (2+3) = 12/5 = 2.4 km/h. Total distance = 2.4 * 3 = 7.2 km.

Let d be the distance covered at 40 km/hr in time t. The remaining distance (55-d) is covered at 50 km/hr. Total time is 1.25 hours. So, t + (55-40t)/50 = 1.25. Multiplying by 50: 50t + 55 - 40t = 62.5, which gives 10t = 7.5, so t = 0.75 hours or 45 minutes.

Let distance be D. Time taken at 20 km/hr is D/20, and at 30 km/hr is D/30. The difference in time is 20 minutes (10 minutes late vs 10 minutes early), which is 1/3 hour. Solving D/20 - D/30 = 1/3 gives D/60 = 1/3, so D = 20 km.

Average speed = (2 * d1 * d2) / (d1 + d2) for equal distances. Average speed = (2 * 4 * 3) / (4 + 3) = 24 / 7 = 3.428... which rounds to 3.43 km/hr.

Speed = 54 km/h = 54 * (5/18) m/s = 15 m/s. Time = Distance / Speed = 90 / 15 = 6 seconds.

Let total distance be 2D and initial speed be v. Time taken = D/v + D/v = 2D/v. New scenario: D/v + D/(1.5v) = 2D/v - 2. D/v * (1 + 2/3) = 2D/v - 2. (5/3) * (D/v) = 2D/v - 2. 2 = (1/3) * (D/v) => D/v = 6. Total time = 12 hours. If speed is decreased by 50% in the second half: D/v + D/(0.5v) = 6 + 12 = 18 hours.

Thief speed = 5 m/s. Man speed = 7.5 m/s. Thief starts 10s early, distance = 50m. Man starts 10m from lady, so distance to cover is 50+10 = 60m. Relative speed = 7.5 - 5 = 2.5 m/s. Time to catch = 60 / 2.5 = 24s. Distance covered by man = 24 * 7.5 = 180m.

The student travels 5 km in 40 - 15 = 25 minutes (5/12 hours). Speed of student = 5 / (5/12) = 12 km/hr. Relative speed logic: The number of overtakes depends on the speed difference. With 60 overtakes in 25 minutes, the vehicle stream speed is 60 km/hr.

Convert speed from kmph to m/s: 90 * (5/18) = 25 m/s. Time = Distance / Speed = 150 / 25 = 6 seconds.

Let D be the distance and v be the jogging speed. v = D/4. The bike speed is v/4 = D/16. The bike covers D/8. Time = (D/8) / (D/16) = 2 hours, which is 120 minutes.